您的SQL语法有错误;检查与您的MySQL服务器版本相对应的手册,以便在第4行的')'附近使用正确的语法

问题描述 投票:0回答:1

这是我第一次尝试使用mySQL,并且在尝试将我的jotform php链接到mySQL数据库时遇到了一个问题。

这是我对数据库的查询:

$query = "SELECT * FROM `new` WHERE `submission_id` = '$submission_id'";
$sqlsearch =mysql_query ($query) or die(mysql_error());
$resultcount = mysql_numrows($sqlsearch);

if ($resultcount > 0) {

mysql_query("UPDATE `new` SET 
                            `captainsfull` = '$captainsfull',
                            `email14` = '$email14',
                            `teamcaptains` = '$teamcaptains',
                            `teammate2` = '$teammate2',
                            `teammate3` = '$teammate3',  
            `teammate4` = '$teammate4',     
            `teammate511` = '$teammate511',
                         WHERE `submission_id` = '$submission_id'") 
 or die(mysql_error());

} else {

mysql_query("INSERT INTO `new` (submission_id, formID, IP, 
                                                                      captainsfull, email14, teamcaptains, teammate2, teammate3, teammate4, teammate511) 
                           VALUES ('$submission_id', '$formID', '$ip', 
                                             '$captainsfull', '$email14', '$teamcaptains', '$teammate2', '$teammate3', '$teammate4', '$teammate511') ") 
or die(mysql_error());  

}
php mysql syntax version jotform
1个回答
0
投票

第4行错误消息“near')'中的线索”指向else块中INSERT语句中的问题。

我没有看到任何明显的错误。输出查询字符串以及mysql错误对于调试非常有用。就像是:

$qry = "INSERT INTO `new` 
       (submission_id, formID, IP, captainsfull, email14, teamcaptains, 
        teammate2, teammate3, teammate4, teammate511) 
        VALUES ('$submission_id', '$formID', '$ip', '$captainsfull', '$email14',
        '$teamcaptains', '$teammate2', '$teammate3', '$teammate4', '$teammate511') ";

mysql_query($qry) or die($qry . mysql_error());

(如果可能的话,请考虑从已弃用的mysql函数迁移到mysqli或PDO。)

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